Mysql: Fill column with consecutive numbers of days in a month - mysql

Question

I have this table: "sales"
+-------------+---------+
| date | total |
+-------------+---------+
| 2018-12-04 | 269.10 |
| 2018-12-05 | 29.00 |
| 2018-12-06 | 107.10 |
| 2018-12-06 | 34.00 |
| 2018-12-08 | 69.50 |
| 2018-12-08 | 223.00 |
| 2018-12-08 | 68.00 |
| 2018-12-09 | 99.00 |
| 2018-12-10 | 59.50 |
| ... | ... |
+-------------+---------+
I'm trying this query
SELECT DAY(date) AS Days,
SUM(CASE WHEN MONTH(date) = 12 THEN total ELSE NULL END) AS December
FROM sales WHERE YEAR(date) = 2018 GROUP BY date
And I get
+-------+----------+
| Days | December |
+-------+----------+
| 4 | 269.10 |
| 5 | 29.00 |
| 6 | 141.10 |
| 8 | 360.50 |
| 9 | 99.00 |
| 10 | 59.50 |
| ... | ... |
+-------+----------+
But I want consecutive days like this:
+-------+----------+
| Days | December |
+-------+----------+
| 1 | NULL |
| 2 | NULL |
| 3 | NULL |
| 4 | 269.10 |
| 5 | 29.00 |
| 6 | 141.10 |
| 7 | NULL |
| 8 | 360.50 |
| 9 | 99.00 |
| 10 | 59.50 |
| ... | ... |
| 31 | 123.00 |
+-------+----------+
Can you help me plss..
PS: I have several months and years in "date" column from "sales" table.

Answer 1

This recursive CTE generates a list of dates corresponding to the month and year specified in the doi CTE, and then LEFT JOINs that to the sales table to get the sales for that month. It will work for any month/year, just change the values in the doi CTE, and the title of the SUM column (currently December) to suit.
WITH RECURSIVE doi AS (
SELECT 12 AS month,
2018 AS year
),
cte AS (
SELECT DATE(CONCAT_WS('-', year, month, 1)) AS date
FROM doi
UNION ALL
SELECT date + INTERVAL 1 DAY
FROM cte
WHERE date < LAST_DAY(date)
)
SELECT DAY(cte.date) AS Days,
ROUND(SUM(s.total),2) AS December
FROM cte
LEFT JOIN sales s ON s.date = cte.date
GROUP BY cte.date
ORDER BY cte.date
Output is too long to show here but can be seen at this demo on dbfiddle
Update
To expand this query to cover an entire year requires changing the approach slightly in terms of generating an entire year's worth of dates, and then using conditional aggregation to get the sums for each day of each month:
WITH RECURSIVE doi AS (
SELECT 2018 AS year
),
cte AS (
SELECT DATE(CONCAT_WS('-', year, 1, 1)) AS date
FROM doi
UNION ALL
SELECT date + INTERVAL 1 DAY
FROM cte
CROSS JOIN doi
WHERE date < DATE(CONCAT_WS('-', doi.year, 12, 31))
)
SELECT DAY(cte.date) AS Days,
ROUND(SUM(CASE WHEN MONTH(s.date) = 1 THEN s.total END),2) AS January,
ROUND(SUM(CASE WHEN MONTH(s.date) = 2 THEN s.total END),2) AS February,
ROUND(SUM(CASE WHEN MONTH(s.date) = 3 THEN s.total END),2) AS March,
ROUND(SUM(CASE WHEN MONTH(s.date) = 4 THEN s.total END),2) AS April,
ROUND(SUM(CASE WHEN MONTH(s.date) = 5 THEN s.total END),2) AS May,
ROUND(SUM(CASE WHEN MONTH(s.date) = 6 THEN s.total END),2) AS June,
ROUND(SUM(CASE WHEN MONTH(s.date) = 7 THEN s.total END),2) AS July,
ROUND(SUM(CASE WHEN MONTH(s.date) = 8 THEN s.total END),2) AS August,
ROUND(SUM(CASE WHEN MONTH(s.date) = 9 THEN s.total END),2) AS September,
ROUND(SUM(CASE WHEN MONTH(s.date) = 10 THEN s.total END),2) AS October,
ROUND(SUM(CASE WHEN MONTH(s.date) = 11 THEN s.total END),2) AS November,
ROUND(SUM(CASE WHEN MONTH(s.date) = 12 THEN s.total END),2) AS December
FROM cte
LEFT JOIN sales s ON s.date = cte.date
GROUP BY DAY(cte.date)
ORDER BY DAY(cte.date)
Demo on dbfiddle

Answer 2

generate your months using union and do right join
select t1.d as Days
, sum(iif(month(date) = 12, total, null) as December
from sales
right join (select 1 as d
union select 2 union select 3 union select 4 union select 5 union select 6
union select 7 union select 8 union select 9 union select 10 union select 11
.... ) as t1 on t1.d = day(date)
where year(date) = 2012
group by date
if you are using mysql v8.0, you can use recursive queries.
with recursive cte as(
select 1 as d
union all
select d + 1 from cte where d < day(last_day('2019-12-01'))
)
select coalesce(day(s.date), t1.d) as Days
, sum(iif(month(s.date) = 12, total, null) as December
from sales s
right join cte as t1 on t1.d = day(s.date)
where year(date) = 2012
group by coalesce(day(s.date), t1.d)